Question

 if theta + phi = alpha and tan theta=x tan phi, then prove that sin(theta-phi)=k-1/k+1 sin alpha




please answer this question its urgent :(​

Answer 1

Answer:

$\implies\:sin(\theta-\phi)=(\frac{k-1}{k+1})sin\alpha$

Step-by-step explanation:

Formula used:

$sin(A+B)=sinA\:cosB+cosA\:sinB$

$sin(A-B)=sinA\:cosB-cosA\:sinB$

Given:

$\theta+\phi=\alpha$...........(1)

and

$tan\theta=k\:tan\phi$

$\implies\:\frac{sin\theta}{cos\theta}=k\:\frac{sin\phi}{cos\phi}$

$sin\theta\:cos\phi=k\:cos\theta\:sin\phi$

Now,

$sin(\theta-\phi)$

$=k\:cos\theta\:sin\phi-cos\theta\:sin\phi$

$=(k-1)\:cos\theta\:sin\phi$

$=\frac{(k-1)}{2}[2\:cos\theta\:sin\phi]$

$=\frac{(k-1)}{2}[sin(\theta+\phi)-sin(\theta-\phi)]$

$=\frac{(k-1)}{2}[sin\alpha-sin(\theta-\phi)]$

$\implies\:sin(\theta-\phi)=\frac{(k-1)}{2}sin\alpha-\frac{(k-1)}{2}sin(\theta-\phi)$

$\implies\:sin(\theta-\phi)+\frac{(k-1)}{2}sin(\theta-\phi)=\frac{(k-1)}{2}sin\alpha$

$\implies\:sin(\theta-\phi)(\frac{2+(k-1)}{2})=\frac{(k-1)}{2}sin\alpha$

$\implies\:sin(\theta-\phi)(\frac{k+1}{2})=\frac{(k-1)}{2}sin\alpha$

$\implies\:sin(\theta-\phi)(k+1)=(k-1)sin\alpha$

$\implies\:sin(\theta-\phi)=(\frac{k-1}{k+1})sin\alpha$